The answer contains such elements $$$a_i$$$ that $$$a_i+1=1$$$ since Chutki starts counting $$$1,2,3...$$$ again. Also add to the answer the last element $$$a_n$$$.

#include <bits/stdc++.h>
#define int long long
#define lop(i,a,b,c) for (int i=a;i<b;i+=c)
#define rlop(i,a,b,c) for (int i=a-1;i>=b;i-=c)
#define prii pair <int,int>
#define PB push_back
#define S second
#define F first
#define B begin()
#define E end()
using namespace std;

/*......................................................................*/

void solve(){
	int n;cin>>n;
	int arr[n+1];vector <int> vec;int p=0;
	lop(i,0,n,1){
		cin>>arr[i];
		if (arr[i]>p)p=arr[i];
		else {
			vec.PB(p);p=1;
		}
	}vec.PB(p);
	cout<<vec.size()<<"\n";
	lop (i,0,vec.size(),1)cout<<vec[i]<<" ";
}
int32_t main(){
	int t;t=1;
	//cin>>t;
	while (t--){
		solve();
	}
	return 0;
}

n=int(input())
a=list(map(int,input().split()))
ans=[]
for i in range(n-1):
	if a[i+1]==1:
		ans.append(a[i])
ans.append(a[-1])
print(len(ans))
print(*ans)

The XOR of two consecutive numbers will always be of the form $$$2^{x}-1$$$(trivial) for some $$$x>=0$$$
So we can convert the given number $$$n$$$ in this form for some value $$$x$$$
Thus, $$$2^{x}+1=n$$$
$$$x$$$ will be equal to $$$log_2(n+1)$$$
So now as we require the minimum possible value of $$$M$$$, it will be $$$(2^{x-1}-1)$$$.
The answer will be $$$-1$$$ for $$$n$$$ which is not of this form $$$2^{x}-1$$$

from math import log2
for _ in range(int(input())):
	n=int(input())
	k=log2(n+1)
	if k%1==0:
		k=int(k)
		print(2**(k-1)-1)
	else:
		print(-1)

To approach this question you should know about divisor properties
Only perfect square have the property to have odd number of divisor {why ? Think about it}
so the condition $$$(p+q)\%2=0$$$ to be satisfied one number should be a perfect square and other should be a non-perfect square. Now you need to minimize the difference between those two numbers. To minimize the difference between those two number you should look for a number nearby half of the given number. {In practice look for the perfect square less than $$$n/2$$$ and perfect square greater then $$$n/2$$$ where $$$n$$$ is the given number, now check which have minimum difference keeping in mind only one number can be a perfect square i.e., both the numbers should not be a perfect square}.

#include <bits/stdc++.h>
#define int long long
#define lop(i,a,b,c) for (int i=a;i<b;i+=c)
#define rlop(i,a,b,c) for (int i=a-1;i>=b;i-=c)
#define prii pair <int,int>
#define PB push_back
#define S second
#define F first
#define B begin()
#define E end()
using namespace std;

/*......................................................................*/

void solve(){
	int n;cin>>n;
	int ans=1e18;int a=-1,b=-1;

	int r=sqrt(n/2);
		lop (i,-1,2,1){
			int re=r+i;
			int df=n-re*re;//cout<<re<<" "<<i<<" "<<df<<"\n";
			int z=sqrt(df);
			if (z*z==df or df<0 or re<=0)continue;
			//cout<<re<<" "<<df<<" "<<abs(re*re-df)<<"\n";
			if (ans>abs(re*re-df)){
				ans=abs(re*re-df);
				a=re*re;b=df;
			}
		}
	
	if (a==-1){
		cout<<a<<"\n";return;
	}
	else {
		cout<<min(a,b)<<" "<<max(a,b)<<"\n";
	}
}
int32_t main(){
	int t;
	cin>>t;
	while (t--){
		solve();
	}
	return 0;
}

from math import sqrt,floor,ceil
from sys import stdin,stdout
input=stdin.readline
for _ in range(int(input())):
  a = int(input())
  b=-1;c=-1;curr=10**15
  res=floor(sqrt(a//2))
  for i in range(-1,2):
    new=res+i
    diff=a-new**2
    if diff>0 and sqrt(diff)%1==0:
      continue
    if curr>abs(diff-new**2):
      curr=abs(diff-new**2)
      c=diff;b=new**2
  if b>0 and c>0:
    print(min(b,c),max(b,c))
  else:
    print(-1)

We will find the value of K by finding its separate bits in binary representation.
Maximum number of bits present in $$$K$$$ can be 40 {think why!}.
For $$$i^{th}$$$ bit, if $$$x$$$ elements have this bit ON, implies $$$N - x$$$ elements have this bit OFF.
-> If this bit is ON in $$$K$$$ , then its contribution in final sum is $$$(N-x)*2^{i}$$$
-> If this bit is OFF in $$$K$$$, then its contribution in final sum is $$$x*2^{i}$$$
If $$$(N-x)>x$$$, then this bit will be OFF in K , otherwise this bit will be ON in K.

Hence, greedily we can find the value of K.

from sys import stdin,stdout
input=stdin.readline
for _ in range(int(input())):
	n = int(input())
	a = list(map(int, input().split()))
	bits = [0 for i in range(40)]
	for x in a:
		for j in range(40):
			bits[j]+=(x>>j & 1)
	ans=0
	for i in range(40):
		if (n-bits[i])<bits[i]:
			ans+=(1<<i)
	stdout.write(str(ans)+'\n')

#include <bits/stdc++.h>
#define int long long
#define lop(i,a,b,c) for (int i=a;i<b;i+=c)
#define rlop(i,a,b,c) for (int i=a-1;i>=b;i-=c)
#define prii pair <int,int>
#define PB push_back
#define S second
#define F first
#define B begin()
#define E end()
using namespace std;

/*......................................................................*/

void solve(){
	int n;cin>>n;
	int arr[64];memset(arr,0,sizeof(arr));
	lop (i,0,n,1){
		int a;scanf("%lld",&a);
		string st=bitset<64>(a).to_string();int sz=st.size();
		rlop (j,sz,0,1){
			if (st[j]=='1'){
				arr[sz-1-j]++;
			}
		}
	}
	int ans=0;int r=1;
	lop (i,0,64,1){
		int p=arr[i];
		if (2*p>n){
			ans+=r;
		}
		r=2*r;
	}
	cout<<ans<<"\n";
}
int32_t main(){
	int t;
	cin>>t;
	while (t--){
		solve();
	}
	return 0;
}